I will address the case $N=2$.

The best strategy is the boring one of distributing your energy evenly. This can be proved inductively in the number of games left. If the players need $a$ and $b$ games to win, respectively, then we say the number of games left is $a+b-1$.

It is trivial for $1$ game left, and a simple calculation with $2$ games left, where one side needs to win both games and the other side needs to win either.

Suppose it is true for $g-1$ games, and let there be $g$ games left. Suppose your opponent distributes his/her energy evenly. Consider the strategy of using $x/g + \delta$ energy on the next game. By induction, regardless of the result, the correct strategy for the last $g-1$ games is to distribute your energy evenly. Choosing $x/g + \delta$ energy for the first game and $x/g - \delta/(g-1)$ equity for the remaining games is equivalent to choosing $x/g + \delta$ energy for the last game and $x/g - \delta/(g-1)$ for the first $g-1$ games. But by induction, after the first game where you use the equity $x/g - \delta/(g-1)$, your equity is less than or equal to the choice of distributing your energy evenly among the last $g-1$. Thus, for any $x/g + \delta$, the equity of that choice is less than or equal to the choice of $x/g + \delta(\frac{-1}{g-1})^n$. By the continuity of winning chances and using $|\frac{-1}{g-1}| \lt 1$, the equity using $x/g+\delta$ on the first choice is at most the equity of using $x/g$. So, by induction, it is optimal to distribute your energy equally.

In the first game of a best-of-seven series, spend $1/7$ of your energy.